91Ó°ÊÓ

Given to calculate each of the following integral:

$$\begin{gathered} a.{∫}_0^{10}4\mathrm{π}\left(11.2+0.072{\mathrm{x}}^2\right)dx \\ b.{∫}_0^4\left(62.4\mathrm{π}\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy \\ c.{∫}_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy \\ d.\mathrm{π}{∫}_0^{100}\left(0.75x+250\right)\left(200-\mathrm{x}\right)dx \\ e.\mathrm{π}{∫}_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}y\right)dy \end{gathered}$$

Solving the integral:

$$\begin{gathered} {∫}_0^{10}4\mathrm{π}\left(11.2+0.072{\mathrm{x}}^2\right)dx=4\mathrm{π}{∫}_0^{10}\left(11.2+0.072{\mathrm{x}}^2\right)dx \\ {∫}_0^{10}4\mathrm{π}\left(11.2+0.072{\mathrm{x}}^2\right)dx=\left(4\mathrm{π}\right){\left(11.2x+\frac{0.072{\mathrm{x}}^3}{3}\right)}_0^{10} \\ {∫}_0^{10}4\mathrm{π}\left(11.2+0.072{\mathrm{x}}^2\right)dx=\left(4\mathrm{π}\right)\left(\left(11.2\left(10\right)+\frac{0.072{\left(10\right)}^3}{3}\right)-\left(11.2\left(0\right)+\frac{0.072{\left(0\right)}^3}{3}\right)\right) \\ {∫}_0^{10}4\mathrm{π}\left(11.2+0.072{\mathrm{x}}^2\right)dx=\left(4\mathrm{π}\right)\left(\left(112+24\right)-\left(0+0\right)\right) \\ {∫}_0^{10}4\mathrm{π}\left(11.2+0.072{\mathrm{x}}^2\right)dx=\left(4\mathrm{π}\right)\left(136\right) \\ {∫}_0^{10}4\mathrm{π}\left(11.2+0.072{\mathrm{x}}^2\right)dx=544\mathrm{π} \end{gathered}$$

Solving the integral:

$$\begin{gathered} {∫}_0^4\left(62.4\mathrm{π}\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=\left(62.4\mathrm{π}\right){\left(3.5\right)}^2{∫}_0^4\left(4-\mathrm{y}\right)dy \\ {∫}_0^4\left(62.4\mathrm{π}\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=\left(62.4\mathrm{π}\right){\left(3.5\right)}^2{\left(4x-\frac{{\mathrm{y}}^2}{2}\right)}_0^4 \\ {∫}_0^4\left(62.4\mathrm{π}\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=\left(62.4\mathrm{π}\right){\left(3.5\right)}^2\left(\left(4\left(4\right)-\frac{{\left(4\right)}^2}{2}\right)-\left(4\left(0\right)-\frac{{\left(0\right)}^2}{2}\right)\right) \\ {∫}_0^4\left(62.4\mathrm{π}\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=\left(62.4\mathrm{π}\right)\left(12.25\right)\left(\left(16-8\right)-\left(0-0\right)\right) \\ {∫}_0^4\left(62.4\mathrm{π}\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=\left(62.4\mathrm{π}\right)\left(12.25\right)\left(8\right) \\ {∫}_0^4\left(62.4\mathrm{π}\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=6115.2\mathrm{π} \end{gathered}$$

Solving the integral:

$$\begin{gathered} {∫}_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=62.4{∫}_0^4\left(80+10y-\frac{5}{2}{y}^2\right)dy \\ {∫}_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=62.4{\left(80y+\frac{{10y}^2}{2}-\frac{5y^3}{6}\right)}_0^4 \\ {∫}_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=62.4\left(\left(80\left(4\right)+\frac{{10\left(4\right)}^2}{2}-\frac{5{\left(4\right)}^3}{6}\right)-\left(80\left(0\right)+\frac{{10\left(0\right)}^2}{2}-\frac{5{\left(0\right)}^3}{6}\right)\right) \\ {∫}_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=62.4\left(\left(320+80-\frac{160}{3}\right)-\left(0+0-0\right)\right) \\ {∫}_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=62.4\left(\frac{1040}{3}\right) \\ {∫}_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=21632 \end{gathered}$$

Solving the integral:

$$\begin{gathered} \mathrm{π}{∫}_0^{100}\left(0.75x+250\right)\left(200-\mathrm{x}\right)dx=\mathrm{π}{∫}_0^{100}\left(50000-100\mathrm{x}-0.75{\mathrm{x}}^2\right)dx \\ \mathrm{π}{∫}_0^{100}\left(0.75x+250\right)\left(200-\mathrm{x}\right)dx=\mathrm{π}{\left(50000x-50{\mathrm{x}}^2-0.25{\mathrm{x}}^3\right)}_0^{100} \\ \mathrm{π}{∫}_0^{100}\left(0.75\mathrm{x}+250\right)\left(200-\mathrm{x}\right)d\mathrm{x}=\mathrm{π}\left(\left(50000\left(100\right)-50{\left(100\right)}^2-0.25{\left(100\right)}^3\right)-\left(50000\left(0\right)-50{\left(0\right)}^2-0.25{\left(0\right)}^3\right)\right) \\ \mathrm{π}{∫}_0^{100}\left(0.75\mathrm{x}+250\right)\left(200-\mathrm{x}\right)d\mathrm{x}=\mathrm{π}\left(\left(5000000-500000-250000\right)-\left(0-0-0\right)\right) \\ \mathrm{π}{∫}_0^{100}\left(0.75\mathrm{x}+250\right)\left(200-\mathrm{x}\right)d\mathrm{x}=4250000\mathrm{π} \end{gathered}$$

Solving the integral:

$$\begin{gathered} \mathrm{π}{∫}_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}y\right)dy=\mathrm{π}{∫}_0^6\left(25{\mathrm{y}}^2-214\mathrm{y}+112\right)dy \\ \mathrm{π}{∫}_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}y\right)dy=\mathrm{π}{\left(\frac{25{\mathrm{y}}^3}{3}-107{\mathrm{y}}^2+112x\right)}_0^6 \\ \mathrm{π}{∫}_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}\mathrm{y}\right)d\mathrm{y}=\mathrm{π}\left(\left(\frac{25{\left(6\right)}^3}{3}-107{\left(6\right)}^2+112\left(6\right)\right)-\left(\frac{25{\left(0\right)}^3}{3}-107{\left(0\right)}^2+112\left(0\right)\right)\right) \\ \mathrm{π}{∫}_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}\mathrm{y}\right)d\mathrm{y}=\mathrm{π}\left(\left(1800-3852+672\right)-\left(0-0+0\right)\right) \\ \mathrm{π}{∫}_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}\mathrm{y}\right)d\mathrm{y}=-1380\mathrm{π} \end{gathered}$$

The integrals of the given expressions are:

$$\begin{gathered} {∫}_0^{10}4\mathrm{π}\left(11.2+0.072{\mathrm{x}}^2\right)dx=544\mathrm{π} \\ {∫}_0^4\left(62.4\mathrm{π}\right){\left(3.5\right)}^2\left(4-\mathrm{y}\right)dy=6115.2\mathrm{π} \\ {∫}_0^{4}62.4\left(10+\frac{5}{2}y\right)\left(8-y\right)dy=21632 \\ \mathrm{π}{∫}_0^{100}\left(0.75\mathrm{x}+250\right)\left(200-\mathrm{x}\right)d\mathrm{x}=4250000\mathrm{π} \\ \mathrm{π}{∫}_0^6\left(-2\mathrm{y}+1.12\right)\left(100-\frac{25}{2}\mathrm{y}\right)d\mathrm{y}=-1380\mathrm{π} \end{gathered}$$